tag:blogger.com,1999:blog-4236175346699186251Fri, 10 Mar 2017 14:51:52 +0000Greenwich maths challenge cryptography cipherMathematics history piProbability freak lotteryPuzzles Eastaway socksScience Museum Moniac false memoriesStatistics sex fallacy paradoxTom Lehrer maths songsWii maths mechanics Newton dynamicsmathematics sex chat-upmaths discworld Ian Stewartnovel mathematics hardy ramanujan littlewoodpi bshm mnemonicsimon singh equations british science festivalGreenwich Maths TimeAn obsolete blog for members of Greenwich University's Department of Mathematical Sciences. This blog ceased in 2011.http://greenwich-maths-time.blogspot.com/noreply@blogger.com (Tony)Blogger73125tag:blogger.com,1999:blog-4236175346699186251.post-7565338805951881287Fri, 10 Mar 2017 14:48:00 +00002017-03-10T14:51:52.488+00:00Greenwich Maths TIme - Festival of MathematicsThe University of Greenwich Maths Centre is proud to be hosting Greenwich Maths Time, the 2017 IMA Festival of Mathematics and its Applications.<br /><br /><span style="font-size: x-large;"><a href="http://www.gre.ac.uk/mathsfestival">Click here to go to the Festival website.</a></span><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />http://greenwich-maths-time.blogspot.com/2017/03/greenwich-maths-time-festival-of.htmlnoreply@blogger.com (Tony)0tag:blogger.com,1999:blog-4236175346699186251.post-8466081055555153469Sun, 29 May 2011 18:00:00 +00002011-05-29T18:58:03.334+01:00Greenwich Maths Challenge 6<a href="http://4.bp.blogspot.com/-fHlDk9bL5n8/TbfE9jGM0xI/AAAAAAAAAFk/NJpvrAZWNAA/s1600/GMC6.jpg"><img style="TEXT-ALIGN: center; MARGIN: 0px auto 10px; WIDTH: 151px; DISPLAY: block; HEIGHT: 67px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5600161223330812690" border="0" alt="" src="http://4.bp.blogspot.com/-fHlDk9bL5n8/TbfE9jGM0xI/AAAAAAAAAFk/NJpvrAZWNAA/s320/GMC6.jpg" /></a><br /><br /><br /><p>Here is the sixth Greenwich Maths Challenge. As usual there will be a small prize for the first correct solution emailed to <a href="mailto:A.Mann@gre.ac.uk">A.Mann@gre.ac.uk</a> by a Greenwich undergraduate.</p><br /><br /><p>If you specify the value of y at n different values of x there is a unique polynomial y=p(x) of degree n-1 which takes these values. But suppose I have a polynomial p(x) of unspecified degree, whose coefficients are all non-negative integers. If you give me a value of x I will tell you the value of p(x). What is the minimum number of evaluations you need to make, to be able to identify all the coefficients of my polynomial? To win the prize you must justify your answer.</p>http://greenwich-maths-time.blogspot.com/2011/04/greenwich-maths-challenge-6.htmlnoreply@blogger.com (Tony)0tag:blogger.com,1999:blog-4236175346699186251.post-8471079534264807348Fri, 20 May 2011 14:34:00 +00002011-05-20T15:37:58.129+01:00Greenwich Maths Talent 2011<a href="http://2.bp.blogspot.com/-vaM_zitB5tc/TdZ8FXMoBlI/AAAAAAAAAGk/HScBONM27SA/s1600/MathSoc.jpg"><img id="BLOGGER_PHOTO_ID_5608806817504691794" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 200px; CURSOR: hand; HEIGHT: 145px; TEXT-ALIGN: center" alt="Greenwich MathSoc logo" src="http://2.bp.blogspot.com/-vaM_zitB5tc/TdZ8FXMoBlI/AAAAAAAAAGk/HScBONM27SA/s320/MathSoc.jpg" border="0" /></a>a graduation showcase supported by the <a href="http://www.ima.org.uk/" target="_blank">Institute of Mathematics and its Applications (IMA)</a>.<br /><br />King William Court 315, University of Greenwich, Friday 27 May, 6 - 7:30pm<br /><br />Cheese, wine and soft drinks will be served.<br /><br />This is an opportuty to celebrate the success of graduating maths students and to find out about the exciting projects they have worked on during the final year.<br /><br />The showcase will be opened by Professor Tom Barnes, Deputy Vice-Chancellor.<br /><br />The twitter hashtag is #GMT2011. There is also <a href="http://mathsoc.cms.gre.ac.uk/GMT/">an event website</a>.<br /><br />All welcome - if you are coming, please tell Noel-Ann Bradshaw (<a href="mailto:N.Bradshaw@gre.ac.uk">N.Bradshaw@gre.ac.uk</a>) for catering purposes.http://greenwich-maths-time.blogspot.com/2011/05/greenwich-maths-talent-2011.htmlnoreply@blogger.com (Tony)0tag:blogger.com,1999:blog-4236175346699186251.post-6079694014385904286Wed, 27 Apr 2011 14:18:00 +00002011-04-27T15:39:48.070+01:00Two new books (and three old ones)<div>Biology is a big application area of mathematics at the moment, and here are two new books which I am looking forward to reading this summer:<br /><br /><img id="BLOGGER_PHOTO_ID_5600268207804421330" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 300px; CURSOR: hand; HEIGHT: 300px; TEXT-ALIGN: center" alt="Ian Stewart, Mathematics of Life" src="http://2.bp.blogspot.com/-JDoI6zZ94Oc/TbgmQ3UYlNI/AAAAAAAAAF0/XD31FXNmNO0/s320/Stewart.jpg" border="0" /><img id="BLOGGER_PHOTO_ID_5600268205252690258" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 300px; CURSOR: hand; HEIGHT: 300px; TEXT-ALIGN: center" alt="Martin Nowak, Super Cooperators" src="http://1.bp.blogspot.com/-5HyQTYLkgGs/TbgmQt0AQVI/AAAAAAAAAFs/IMeoSE0YgLo/s320/Nowak.jpg" border="0" /><br />Stewart says that<br /><br />"Mathematical theory and practice have always gone hand in hand, from the time primitive humans scratched marks on bones to record the phases of the Moon to the current search for the Higgs boson using the Large Hadron Collider. Isaac Newton's calculus informed us about the heavens, and over the past three centuries its successors have opened up the whole of mathematical physics: heat, light, sound, fluid mechanics, and later relativity and quantum theory. Mathematical thinking has become the central paradigm of the physical sciences.<br /><br />"Until very recently, the life sciences were different. There, mathematics was at best a servant. It was used to perform routine calculations and to test the significance of statistical patterns in data. It didn't contribute much conceptual insight or understanding. Most of the time, it might as well not have existed.<br /><br />"Today, this picture is changing. Modern discoveries in biology have opened up a host of important questions, and many of them are unlikely to be answered without significant mathematical input, The variety of matheamtical ideas now being used in the life sciences is enormous, and the demands of biology are stimulating the creation of entirely new mathematics, specifically aimed at living processes. Today's mathematicians and biologists are working together on some of the most difficult scientific problems that the human race has ever tackled - including the nature and origin of life itself.<br /><br />"Biology will be the great mathematical frontier of the twenty-first century." <br /><br />Curiously enough, three of the most exciting maths books I have come across, and which have influenced and inspired my teaching, as students will have noticed, also relate to the mathematics of life. </div><br /><a href="http://4.bp.blogspot.com/-pKRJAWqbPWs/TbgqTIzNkrI/AAAAAAAAAGc/XdmiOuVYs_E/s1600/Axelrod.jpg"><img style="cursor:pointer; cursor:hand;width: 166px; height: 257px;" src="http://4.bp.blogspot.com/-pKRJAWqbPWs/TbgqTIzNkrI/AAAAAAAAAGc/XdmiOuVYs_E/s320/Axelrod.jpg" border="0" alt="Robert Axelrod, The Evolution of Co-operation"id="BLOGGER_PHOTO_ID_5600272644903375538" /></a><br /><a href="http://1.bp.blogspot.com/-fSZ3_kXGlLw/TbgqS_H_xzI/AAAAAAAAAGU/BFQkdQWSI1Q/s1600/Thompson.jpg"><img style="cursor:pointer; cursor:hand;width: 159px; height: 260px;" src="http://1.bp.blogspot.com/-fSZ3_kXGlLw/TbgqS_H_xzI/AAAAAAAAAGU/BFQkdQWSI1Q/s320/Thompson.jpg" border="0" alt="D'ARcy Thopmpson, On Growth and Form"id="BLOGGER_PHOTO_ID_5600272642306197298" /></a><br /><a href="http://3.bp.blogspot.com/-_FINhH3cjkc/TbgqSTbT-4I/AAAAAAAAAGM/gbVudY2fOTw/s1600/41zCe0ju76L__BO2%252C204%252C203%252C200_PIsitb-sticker-arrow-click%252CTopRight%252C35%252C-76_AA300_SH20_OU02_.jpg"><img style="cursor:pointer; cursor:hand;width: 181px; height: 257px;" src="http://3.bp.blogspot.com/-_FINhH3cjkc/TbgqSTbT-4I/AAAAAAAAAGM/gbVudY2fOTw/s320/41zCe0ju76L__BO2%252C204%252C203%252C200_PIsitb-sticker-arrow-click%252CTopRight%252C35%252C-76_AA300_SH20_OU02_.jpg" border="0" alt="Matt Ridley, The Origins of Virtue"id="BLOGGER_PHOTO_ID_5600272630576053122" /></a>http://greenwich-maths-time.blogspot.com/2011/04/two-new-books-and-three-old-ones.htmlnoreply@blogger.com (Tony)0tag:blogger.com,1999:blog-4236175346699186251.post-8099921076959301983Wed, 27 Apr 2011 07:12:00 +00002011-04-27T08:18:54.535+01:00Big Noise<a href="http://1.bp.blogspot.com/-l0c-q8tmAJk/TbfDNnlJitI/AAAAAAAAAFc/2H_VlriFQ1Y/s1600/a_codebook4.jpg"><img id="BLOGGER_PHOTO_ID_5600159300389014226" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 160px; CURSOR: hand; HEIGHT: 143px; TEXT-ALIGN: center" alt="Simon singh's Enigma machine simulator" src="http://1.bp.blogspot.com/-l0c-q8tmAJk/TbfDNnlJitI/AAAAAAAAAFc/2H_VlriFQ1Y/s320/a_codebook4.jpg" border="0" /></a><br /><br /><div>No, not one of our first year lectures, but the Bexley Big Noise STEM Careers Fair. This was an event for year 9 and 10 students in Bexley to promote careers in science, engineering, technology and maths. Kevin and I went with two of our students (thanks Ameli and James) , and we had a wonderful time, demonstrating code-breaking, mathematical modelling and actuarial mathematics, to lively and enthusiastic mathematicians of the future.</div>http://greenwich-maths-time.blogspot.com/2011/04/big-noise.htmlnoreply@blogger.com (Tony)2tag:blogger.com,1999:blog-4236175346699186251.post-3664465955529456495Sat, 09 Apr 2011 16:45:00 +00002011-04-09T18:02:13.393+01:00Young Operational Research Society Conference<a href="http://4.bp.blogspot.com/-e0oJmCTHrxA/TaCRC4ZaWhI/AAAAAAAAADU/ztq5DgV9128/s1600/OR_logo.gif"><img style="MARGIN: 0px 10px 10px 0px; WIDTH: 121px; FLOAT: left; HEIGHT: 73px; CURSOR: hand" id="BLOGGER_PHOTO_ID_5593630215878892050" border="0" alt="" src="http://4.bp.blogspot.com/-e0oJmCTHrxA/TaCRC4ZaWhI/AAAAAAAAADU/ztq5DgV9128/s200/OR_logo.gif" /></a> <br /><div>Last Monday I travelled to Nottingham to the <strong>Young Operational Research Society Conference</strong>. <br /><p>My main reason for attending was to present a paper on my research concerning a new Multi-Objective Evolutionary Algorithm for Portfolio Optimisation. This took place on the Tuesday Morning and was well received. <br /><p>However of even greater benefit was the information I picked up from those working in OR about the skills they needed graduates to have. Nowadays it is important that, as well as having good technical skills, graduates also need good interpersonal skills and business awareness. Much of this can be acquired through the group projects we set at Greenwich and our focus, in the second year, on employability. <br /><p>Many of the people I spoke to were involved in building simulation models. Something I will emphasise when I teach this next year. I took part in a demonstration of a software called SIMUL8 (we will be using this in OR next year). We had to simulate a nightclub and despite the fact that I have never been to one and do not know what Indie music is our team won! <br /><p>More information about the OR Society can be obtained from: http://www.orsoc.org.uk/orshop/(f3aerozzof4l3o45irqjymax)/orhomepage2.aspx</p></div>http://greenwich-maths-time.blogspot.com/2011/04/young-operational-research-society.htmlnoreply@blogger.com (Noel-Ann)0tag:blogger.com,1999:blog-4236175346699186251.post-5830107354562057327Fri, 18 Mar 2011 15:25:00 +00002011-03-18T16:10:56.256+00:00MathSoc's Pi Day Party<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/-ulZVlIStsVk/TYOBUZktL4I/AAAAAAAAADM/hsUioxA6or8/s1600/2cakePies.jpg"><img id="BLOGGER_PHOTO_ID_5585450150331756418" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 200px; CURSOR: hand; HEIGHT: 134px; TEXT-ALIGN: center" alt="" src="http://4.bp.blogspot.com/-ulZVlIStsVk/TYOBUZktL4I/AAAAAAAAADM/hsUioxA6or8/s200/2cakePies.jpg" border="0" /></a><br /><br /><div><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/-XC1ABxauOS0/TYOBT6PWK5I/AAAAAAAAADE/GQlT67tQgZA/s1600/Noel-annPie.jpg"><img id="BLOGGER_PHOTO_ID_5585450141920668562" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 200px; CURSOR: hand; HEIGHT: 134px; TEXT-ALIGN: center" alt="" src="http://3.bp.blogspot.com/-XC1ABxauOS0/TYOBT6PWK5I/AAAAAAAAADE/GQlT67tQgZA/s200/Noel-annPie.jpg" border="0" /></a><br /><br /><br /><br /><div>On 14th March (3.14) the maths staff and students celebrated pi day. We had a short talk on the history of pi which included learning some fun mnemonics like 'How I Need A Drink Alcoholic Of Course' (3.1415926) and having a competition to see who could memorise the most digits of pi. This was won by Nic Mortimer (3rd year - BSc Decision Science) who memorised 101. Karen Richardson from the library came a close second. <p></p></div><div></div><div>We also listened to a pi song - available here: <a href="http://bit.ly/49CFj">http://bit.ly/49CFj</a></div><div><p></p></div><div>This was followed by some pi pie and pi cake made by Noel-Ann Bradshaw, Ameli Gottstein (3rd year - BSc Maths) and James Howe (3rd year - BSc Maths). Pictures by Michael Dullaway (3rd year - BSc Maths).</div></div>http://greenwich-maths-time.blogspot.com/2011/03/mathsocs-pi-day-party.htmlnoreply@blogger.com (Noel-Ann)1tag:blogger.com,1999:blog-4236175346699186251.post-7033880203775112304Fri, 18 Mar 2011 10:34:00 +00002011-03-18T11:02:14.082+00:00Greenwich Christmas QuizCongratulations to Ameli Gottstein who won this year's <a href="http://greenwich-maths-time.blogspot.com/2010/12/greenwich-christmas-maths-challenge.html">Greenwich Christmas Maths Challenge</a>. This completes a clean sweep for Ameli who has won in each of her three years as a Greenwich undergraduate!<br /><br />Here are solutions to the questions.<br /><br />Q1. Identify the mathematicians whose names are given below as anagrams. Accents and punctuation marks such as hyphens are omitted, and spellings are taken from the MacTutor History of Mathematics website <a href="http://www-groups.dcs.st-and.ac.uk/~history/">http://www-groups.dcs.st-and.ac.uk/~history/</a><br /><br />a) ENRW<br />b) EIKLNV<br />c) EEKLPR<br />d) AAIIJNX<br />e) AHILMNOT<br />f) BEILLNORU<br />g) ACDEEHIMRS<br />h) AEMNNNNOUV<br />i) GKLMOOOORV<br />j) AAAEKKLOSVVYQ<br /><br />Answers: (a) Wren, (b) Kelvin, (c) Kepler, (d) Jia Xian, (e) Hamilton, (f) Bernoulli, (g) Archimedes, (h) Von Neumann, (i) Kolmogorov, (j) Kovalevskaya<br /><br />For information on these mathematicians see the MacTutor website.<br /><br />2. A ship is at anchor in a harbour. A spectator sees a ladder dangling from her deck. The bottom four rungs of the ladder are submerged, each rung is two centimetres wide and the rungs are eleven centimetres apart. The tide is rising by eighteen centimetres per hour. After two hours, how many rungs will be submerged?<br /><br />Answer: still only four rungs. The ship and ladder rise with the tide!<br /><br />Q3. For Spanish, Russian or Hebrew, it’s 1. For German, it’s 7. For French, it’s 14. What is it for English?<br /><br />Answer: The answer is 7. It’s the first integer whose name in the language has more than one syllable.<br /><br />Q4. How many people is "Twice two pairs of twins"?<br /><br />Answer: Eight<br /><br />Q5. Each integer from 1 to 10^10 (ten billion) is written out in full (for example 211 would be "two hundred eleven" and 1042 would be "one thousand forty two" - the word "and" is not used), and the numbers are then listed in alphabetical order (ignoring spaces and hyphens). What is the first odd number to appear in the list?<br /><br />The book answer is 8,018, 018,885 (eight billion, eighteen million, eighteen thousand, eight hundred eighty five).<br /><br />Q6. The Ruritanian National Library contains more books than any single book on its shelves contains words. No two of its books contain the same number of words. Can you say how many words are contained in one of its books?<br /><br />Answer: At least one book contains zero words. Use the Pigeon-hole Principle!<br /><br />Q7. A maths lecturer has a collection of eighteenth-century mathematical pornography in two bookcases in a room 9 by 12 feet (a foot is an archaic unit of measure about 30cm long). Bookcase AB is 8.5 feet long and bookcase CD is 4.5 feet long. The bookcases are positioned centrally on each wall and are one inch from the wall, as shown in the diagram.<br /><br /><a href="http://3.bp.blogspot.com/-ee-r8xy-idE/TYM6-xhadJI/AAAAAAAAAFU/KwV6sLZPIqY/s1600/bookcases.jpg"><img id="BLOGGER_PHOTO_ID_5585372812989330578" style="FLOAT: left; MARGIN: 0px 10px 10px 0px; WIDTH: 150px; CURSOR: hand; HEIGHT: 320px" alt="Diagram" src="http://3.bp.blogspot.com/-ee-r8xy-idE/TYM6-xhadJI/AAAAAAAAAFU/KwV6sLZPIqY/s320/bookcases.jpg" border="0" /></a><br /><br />Some students are going to visit the lecturer and she wishes to protect the students from the books and vice versa. The lecturer decides that the best way to do this is to turn around the two bookcases so that each is in its starting position but with the ends reversed so that the books are all facing the wall. The bookcases are so heavy that the only way to move them is to keep one end on the floor as a pivot while the other end is swung in a circular arc. The bookcases are so narrow that for the purpose of this problem we can consider them as straight line segments. What is the minimum number of swings required to reverse the two bookcases?<br /><br /><br /><br />(For more about eighteenth-century mathematical pornography, see <a href="http://www.gresham.ac.uk/event.asp?PageId=45&EventId=1028">Patricia Fara's recent Gresham College lecture</a>.)<br /><br /><br />Answer: Eight swings are enough. For example, (1) Swing end B clockwise 90 degrees; (2) swing A clockwise 30 degrees; (3) swing B anti-clockwise 60 degrees; (4) swing A clockwise 30 degrees; (5) swing B clockwise 90 degrees; (6) swing C clockwise 60 degrees; (7) Swing D anti-ckockwise 300 degrees; (8) swing C clockwise 60 degrees.<br /><br />Q8. The number 2 to the power 29 has nine digits, all different: which digit is missing? (Calculator not required.)<br /><br />The answer is 4.<br /><br />The way to do this without a calculator is to use the fact that, if you divide a number by 9, the remainder you get is the same as the remainder when you divide the sum of its digits by 9. (This is the basis of lots of maths tricks you can impress your friends with.) For example, 123456 divided by 9 has remainder 3; so does 1+2+3+4+5+6 = 21.<br /><br />Consider the powers of 2 divided by 9 (draw your own table). It’s easy to see that the pattern repeats every six powers. (You may know Euler’s Theorem relatign to this.) So 229 will have the same remainder divided by 9 as 25 does: this is, 5, which must be the sum of the nine digits. Now the sum of all the ten possible digits 1+2+3+…+9+0 is 45, so the sum of eight of them is between 36 and 45, and the only number in that range which leaves remainder 5 when divided by 9 is 41, which means 4 must be left out.<br /><br />Q9. What is the 99th digit to the right of the decimal point in the decimal expansion of (1+√2) to the power 500? (Calculator not required.) (In case this isn't displayed correctly by your browser, the expression is (1+sqrt(2))^500, where sqrt(2) means the positive square root of 2.<br /><br />The answer is 9.<br /><br />At first sight this seems an impossible difficult question without doing detailed calculations to a huge number of decimal places. But consider the following.<br />Let x = (1+√2) and x’ = (1-√2). What happens when we add powers of x and x’?<br />Let y = (1+√2)^500 + (1-√2)^500<br /><br />We can use the binomial theorem to expand the two powers. Now, for odd powers of √2, the coefficients will have opposite signs in the expansions of x^500 and x’^500 and they will cancel, so we have an expression for y in terms of even powers of √2. But an even power of √2 is a power of 2, so it is an integer. So y is an integer.<br /><br />Now, what is x’^500? Well, x’ = 1-√2 is about -.414: it’s a negative number with absolute value less than ½. So the 500th power of x’ is positive and is less than (1/2)^500, which is very small – about 4x10^-192. So the decimal expansion of x’ begins with more than 100 zeroes.<br /><br />So x plus a positive number beginning with 100 zeroes gives an integer: so the first 99 digits after the decimal point in x must all be 9s.<br /><br />When I first saw this problem I thought it was impossible so I looked at the answer straightaway. But having read the answer it doesn’t seem so impossible – because when one has a term like (1+√2) in an expression it’s a common trick to think about (1-√2). It’s hard to see any other way one could tackle this problem! So having looked at the answer, I’m now sure I could have done it myself if I’d bothered. (I may be being over-optimistic!)<br /><br />Q10. Here are two messages enciphered using substitution ciphers. What do they say?<br /><br />a) PREEZ LAEDHKPFH FSO AFYYZ SRT ZRFE!<br />b) RIFFY SRPWUZQIU! OWURWVM YAE I MAAX RALWXIY!<br /><br />Answers: (a) MERRY CHRISTMAS AND A HAPPY NEW YEAR!<br />(b) HAPPY CHRISTMAS! WISHING YOU A GOOD HOLIDAY!<br /><br />(b) was designed to avoid the letter E to make frequency analysis a little harder.<br /><br />Questions 3, 4, 8 and 9 came from <em>Mathematical Mind-Benders</em> by Peter Winkler and Question 5 from the same author’s <em>Mathematical Puzzles: A Connoisseur’s Collection</em>. Questions 2, 6 and 7 are from David Wells's <em>The Penguin Book of Curious and Interesting Puzzles</em>.http://greenwich-maths-time.blogspot.com/2011/03/greenwich-christmas-quiz.htmlnoreply@blogger.com (Tony)0tag:blogger.com,1999:blog-4236175346699186251.post-709945804704182982Tue, 21 Dec 2010 11:38:00 +00002010-12-21T13:28:59.570+00:00Greenwich Christmas Maths Challenge 2010A small prize will be offered for the best set of solutions emailed to A.Mann@gre.ac.uk by a Greenwich student before 5pm on Monday 10 January 2011. In the event of a tie, a winner will be chosen randomly. The judges' decision is final. For obvious reasons, the source of these questions won’t be revealed until afterwards. The quiz has ten questions of varying degrees of difficulty: some I think are pretty hard! On past experience, you may well not need to answer them all in order to win.<br /><br />Q1. Identify the mathematicians whose names are given below as anagrams. Accents and punctuation marks such as hyphens are omitted, and spellings are taken from the MacTutor History of Mathematics website http://www-groups.dcs.st-and.ac.uk/~history/<br />a) ENRW<br />b) EIKLNV<br />c) EEKLPR<br />d) AAIIJNX<br />e) AHILMNOT<br />f) BEILLNORU<br />g) ACDEEHIMRS<br />h) AEMNNNNOUV<br />i) GKLMOOOORV<br />j) AAAEKKLOSVVY<br /><br />Q2. A ship is at anchor in a harbour. A spectator sees a ladder dangling from her deck. The bottom four rungs of the ladder are submerged, each rung is two centimetres wide and the rungs are eleven centimetres apart. The tide is rising by eighteen centimetres per hour. After two hours, how many rungs will be submerged?<br /><br />Q3. For Spanish, Russian or Hebrew, it’s 1. For German, it’s 7. For French, it’s 14. What is it for English?<br /><br />Q4. How many people is "Twice two pairs of twins"?<br /><br />Q5. Each integer from 1 to 10^10 (ten billion) is written out in full (for example 211 would be "two hundred eleven" and 1042 would be "one thousand forty two" - the word "and" is not used), and the numbers are then listed in alphabetical order (ignoring spaces and hyphens). What is the first odd number to appear in the list?<br /><br />Q6. The Ruritanian National Library contains more books than any single book on its shelves contains words. No two of its books contain the same number of words. Can you say how many words are contained in one of its books?<br /><br />Q7. A maths lecturer has a collection of eighteenth-century mathematical pornography in two bookcases in a room 9 by 12 feet (a foot is an archaic unit of measure about 30cm long). Bookcase AB is 8.5 feet long and bookcase CD is 4.5 feet long. The bookcases are positioned centrally on each wall and are one inch from the wall, as shown in the diagram.<br /><p><img id="BLOGGER_PHOTO_ID_5553100217488495922" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 150px; CURSOR: hand; HEIGHT: 320px; TEXT-ALIGN: center" alt="Diagram for bookcase problem" src="http://3.bp.blogspot.com/_oCutOBezbNs/TRCTOemL1TI/AAAAAAAAAE8/EF5I2B18hao/s320/bookcases.jpg" border="0" /> Some students are going to visit the lecturer and she wishes to protect the students from the books and vice versa. The lecturer decides that the best way to do this is to turn around the two bookcases so that each is in its starting position but with the ends reversed so that the books are all facing the wall. The bookcases are so heavy that the only way to move them is to keep one end on the floor as a pivot while the other end is swung in a circular arc. The bookcases are so narrow that for the purpose of this problem we can consider them as straight line segments. What is the minimum number of swings required to reverse the two bookcases? </p><p>(For more about eighteenth-century mathematical pornography, see <a href="http://www.gresham.ac.uk/event.asp?PageId=45&EventId=1028">Patricia Fara's recent Gresham College lecture</a>.)<br /><br />Q8. The number 2 to the power 29 has nine digits, all different: which digit is missing? (Calculator not required.)<br /><br />Q9. What is the 99th digit to the right of the decimal point in the decimal expansion of (1+√2) to the power 500? (Calculator not required.) (In case this isn't displayed correctly by your browser, the expression is (1+sqrt(2))^500, where sqrt(2) means the positive square root of 2.<br /><br />Q10. Here are two messages enciphered using substitution ciphers. What do they say?<br /><br />a) PREEZ LAEDHKPFH FSO AFYYZ SRT ZRFE!<br />b) RIFFY SRPWUZQIU! OWURWVM YAE I MAAX RALWXIY!</p>http://greenwich-maths-time.blogspot.com/2010/12/greenwich-christmas-maths-challenge.htmlnoreply@blogger.com (Tony)0tag:blogger.com,1999:blog-4236175346699186251.post-1041760260071025323Mon, 06 Dec 2010 21:09:00 +00002010-12-06T21:21:16.674+00:00Greenwich Maths Challenge 5<a href="http://3.bp.blogspot.com/_oCutOBezbNs/TP1Rr8nIYRI/AAAAAAAAAE0/GkxP3-f-RF4/s1600/GMC5.jpg"><img id="BLOGGER_PHOTO_ID_5547680131436798226" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 151px; CURSOR: hand; HEIGHT: 67px; TEXT-ALIGN: center" alt="" src="http://3.bp.blogspot.com/_oCutOBezbNs/TP1Rr8nIYRI/AAAAAAAAAE0/GkxP3-f-RF4/s320/GMC5.jpg" border="0" /></a> The problem posed was as follows:<br /><br />"In the small country of Mathsland, the citizens are obsessed with politics. Each one passionately supports one of the three political parties, which are the Coffee, Milk and Water parties.<br /><br />Whenever two citizens meet, they discuss politics. If they support the same party, they don’t change their allegiance, but if two citizens who support different parties meet, they are both so persuasive that each of them abandons their previous allegiance and supports the third party. Thus if Milk and Water supporters meet, they both change to support Coffee.<br /><br />The repressive laws of Mathsland forbid any gathering of more than two people so all political discussions are limited to the above. If at any time all citizens support a single party, that party will declare a dictatorship and the other two parties will be abolished.<br /><br />Initially there are 13 supporters of Coffee, 15 of Milk and 17 of Water. Find the shortest possible sequence of meetings which results in a dictatorship, or prove that under these conditions no party will ever command the support of every citizen."<br /><br />This was the most popular Greenwich Maths Challenge yet, with entries almost equally divided between those who claimed to have found such a sequence and those who claimed to have proved it was impossible. The latter were correct, with several excellent answers submitted. In the opinion of the judges, the first completely valid proof came from Aaron Lang, who wins the prize.<br /><br />The puzzle is an old one: the traditional scenario involves chameleons but we didn't want it to be too easy to google so we changed the setting. It can be found in several books of puzzles. One solution, given in Terence Tao's <em>Solving Mathematical Problems</em>, is to assign values of 0 to supporters of Coffee, 1 to Milk and 2 to Water. The initial total is 49, which gives a remainder of 1 when divided by 3. One can verify that each of the three possible moves changes the total - for example Coffee meeting Milk removes one of each, subtracting one from the total, but adds two Waters, adding 4 - but that each change does not alter the remainder after division by 3. So at any stage the remainder must remain at 1, but any of the desired solutions needs remainder zero, so they are impossible.http://greenwich-maths-time.blogspot.com/2010/12/greenwich-maths-challenge-5.htmlnoreply@blogger.com (Tony)1tag:blogger.com,1999:blog-4236175346699186251.post-3668063071467150050Sun, 07 Nov 2010 08:55:00 +00002010-11-07T16:35:49.298+00:00Greenwich Maths Challenge 5<a href="http://3.bp.blogspot.com/_oCutOBezbNs/TNZ3lr9EmHI/AAAAAAAAAEs/TXNNGAqmLtE/s1600/GMC5.jpg"><img id="BLOGGER_PHOTO_ID_5536744281236346994" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 151px; CURSOR: hand; HEIGHT: 67px; TEXT-ALIGN: center" alt="GMC5 logo" src="http://3.bp.blogspot.com/_oCutOBezbNs/TNZ3lr9EmHI/AAAAAAAAAEs/TXNNGAqmLtE/s320/GMC5.jpg" border="0" /></a> A small prize will be awarded for the first correct solution sent to Tony Mann (A.Mann@gre.ac.uk) by a Greenwich undergraduate.<br /><br />In the small country of Mathsland, the citizens are obsessed with politics. Each one passionately supports one of the three political parties, which are the Coffee, Milk and Water parties.<br /><br />Whenever two citizens meet, they discuss politics. If they support the same party, they don’t change their allegiance, but if two citizens who support different parties meet, they are both so persuasive that each of them abandons their previous allegiance and supports the third party. Thus if Milk and Water supporters meet, they both change to support Coffee.<br /><br />The repressive laws of Mathsland forbid any gathering of more than two people so all political discussions are limited to the above.<br /><br />If at any time all citizens support a single party, that party will declare a dictatorship and the other two parties will be abolished.<br /><br />Initially there are 13 supporters of Coffee, 15 of Milk and 17 of Water. Find the shortest possible sequence of meetings which results in a dictatorship, or prove that under these conditions no party will ever command the support of every citizen.http://greenwich-maths-time.blogspot.com/2010/11/greenwich-maths-challenge-5.htmlnoreply@blogger.com (Tony)0tag:blogger.com,1999:blog-4236175346699186251.post-7125044043408740810Thu, 04 Nov 2010 08:14:00 +00002010-11-04T08:18:14.408+00:00COMING SOON<a href="http://1.bp.blogspot.com/_oCutOBezbNs/TNJrtqPic9I/AAAAAAAAAEk/_vHpOND3oF0/s1600/GMC5.jpg"><img id="BLOGGER_PHOTO_ID_5535605324169507794" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 151px; CURSOR: hand; HEIGHT: 67px; TEXT-ALIGN: center" alt="GMC5 logo" src="http://1.bp.blogspot.com/_oCutOBezbNs/TNJrtqPic9I/AAAAAAAAAEk/_vHpOND3oF0/s320/GMC5.jpg" border="0" /></a><br />Coming soon - <strong>Greenwich Maths Challenge 5</strong>! A tough maths puzzle with a small prize and a lot of glory for the first correct solution from a Greenwich maths undergraduate. GMC5 will be posted at or soon after 10am on on Sunday morning, 7 November 2010. To practice, why not have a look at GMCs 1-4 posted earlier on this blog.http://greenwich-maths-time.blogspot.com/2010/11/coming-soon.htmlnoreply@blogger.com (Tony)0tag:blogger.com,1999:blog-4236175346699186251.post-3264169575284026483Sun, 26 Sep 2010 14:00:00 +00002010-09-26T15:07:11.477+01:00Maths Puzzle CompetitionThe entries for the freshers' maths puzzle competition have now been marked. There were 4 very close groups with marks between 60 and 55 - Roxy, Lena, Shanka and Tammy but way out in front with 71 points was Emma M's group. <br /><br />Well done to all those who took part. Answers are not being posted but some questions will be gone through in MaTT at some point.http://greenwich-maths-time.blogspot.com/2010/09/maths-puzzle-competition.htmlnoreply@blogger.com (Noel-Ann)0tag:blogger.com,1999:blog-4236175346699186251.post-5591108173847888685Thu, 16 Sep 2010 18:43:00 +00002010-09-16T19:49:10.756+01:00pi bshm mnemonicPi Day at the British Science FestivalPi Day (Twitter #PiHunt) yesterday was a great success, with a brilliant audience whose success with Buffon's Needle surpassed our expectations and forced us to depart from our script. And it gave rise to Peter Rowlett's wonderful post-event pi mnemonic<br /><br />Now I away,<br />I leave enlivened by having great and happy<br />historic, enjoyable, happily unanimous joy<br />at the exciting BSHM #pihunt<br /><br />So no excuse for anyone to use 22/7 ever again.<br /><br />(find more at <a href="http://travelsinamathematicalworld.blogspot.com/2010/09/how-i-wish-i-could-calculate-pihunt.html">http://travelsinamathematicalworld.blogspot.com/2010/09/how-i-wish-i-could-calculate-pihunt.html</a>)http://greenwich-maths-time.blogspot.com/2010/09/pi-day-at-british-science-festival.htmlnoreply@blogger.com (Tony)0tag:blogger.com,1999:blog-4236175346699186251.post-9094943495746539933Tue, 14 Sep 2010 14:02:00 +00002010-09-14T15:09:05.972+01:00Mathematics history piPi-Hunting<a href="http://4.bp.blogspot.com/_oCutOBezbNs/TI-AsFz2fXI/AAAAAAAAAEc/mWecNnghVwA/s1600/Ludolf_van_Ceulen.jpg"><img id="BLOGGER_PHOTO_ID_5516769563514404210" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 263px; CURSOR: hand; HEIGHT: 320px; TEXT-ALIGN: center" alt="Ludolf van Ceulen" src="http://4.bp.blogspot.com/_oCutOBezbNs/TI-AsFz2fXI/AAAAAAAAAEc/mWecNnghVwA/s320/Ludolf_van_Ceulen.jpg" border="0" /></a><br /><div>This is Ludolph van Ceulen, who along with Noel-Ann Bradshaw and Tony Mann of Greenwich and Mark McCartney of Ulster will be presenting an event about the digits of the mathematical constant pi at the <a href="http://www.britishscienceassociation.org/web/BritishScienceFestival/index.htm">British Science Festival</a> on Wednesday 15th September. Robin Wilson will be there too but is unlikely to be around at the same time as Ludolph. A highlight wil be the "Pi Moment" at 3:14pm (why?) To find out more, <a href="http://www.britishscienceassociation.org/forms/festival/events/showevent2.asp?EventID=72">come to the event</a>!</div>http://greenwich-maths-time.blogspot.com/2010/09/pi-hunting.htmlnoreply@blogger.com (Tony)0tag:blogger.com,1999:blog-4236175346699186251.post-3004793729506377824Thu, 24 Jun 2010 08:09:00 +00002010-06-24T09:16:39.941+01:00Greenwich Maths Challenge 4 result<a href="http://3.bp.blogspot.com/_oCutOBezbNs/TCMS0r3PyXI/AAAAAAAAAEM/ICXJan7PXP4/s1600/GMC4.jpg"><img id="BLOGGER_PHOTO_ID_5486249467404339570" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 151px; CURSOR: hand; HEIGHT: 67px; TEXT-ALIGN: center" alt="GMC4 logo" src="http://3.bp.blogspot.com/_oCutOBezbNs/TCMS0r3PyXI/AAAAAAAAAEM/ICXJan7PXP4/s320/GMC4.jpg" border="0" /></a> Congratulations to Mike Wakeling who provided the first correct answer to GMC4. This one attracted a lot of interest with many attempted solutions. Many people thought the problem was insoluble (as I did when I first saw it!) This problem came from the late Martin Gardner, and I think it's a gem. If you haven't tried it yet, <a href="http://greenwich-maths-time.blogspot.com/2010/05/greenwich-maths-challenge-4.html">have a go at the problem</a> before reading the solution below.<br /><br />Here's the solution. You have the given quantities of red, yellow, green and blue paint and have to colour the diagram in four colours so that any two regions with a common boundary are different colours. This is impossible if the colours are red, yellow, green and blue. So you have to mix the blue with an equal quantity of red, so that you now have enough yellow for 24 square metres and enough purple, green and red for 16 square metres each: the problem can now be solved!http://greenwich-maths-time.blogspot.com/2010/06/greenwich-maths-challenge-4-result.htmlnoreply@blogger.com (Tony)2tag:blogger.com,1999:blog-4236175346699186251.post-250100184942232693Sat, 29 May 2010 10:48:00 +00002010-05-29T18:46:38.925+01:00Greenwich Maths Challenge 4<a href="http://2.bp.blogspot.com/_oCutOBezbNs/TADxq_-AWCI/AAAAAAAAAD8/xJbPRXwe0xI/s1600/GMC4.jpg"><img id="BLOGGER_PHOTO_ID_5476642867910826018" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 151px; CURSOR: hand; HEIGHT: 67px; TEXT-ALIGN: center" alt="GMC4 logo" src="http://2.bp.blogspot.com/_oCutOBezbNs/TADxq_-AWCI/AAAAAAAAAD8/xJbPRXwe0xI/s320/GMC4.jpg" border="0" /></a> Here, to give people something to do now that the exams have finished, is the final Greenwich Maths Challenge for the 2009/2010 academic year. A prize will be awarded for the first correct solution emailed to <a href="mailto:A.Mann@gre.ac.uk">A.Mann@gre.ac.uk</a> by a Greenwich maths undergraduate.<br /><br /><br /><br /><a href="http://3.bp.blogspot.com/_oCutOBezbNs/TADy3vJidsI/AAAAAAAAAEE/v7UI2mUQfLo/s1600/4ct+problem.jpg"><img id="BLOGGER_PHOTO_ID_5476644186245723842" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 320px; CURSOR: hand; HEIGHT: 312px; TEXT-ALIGN: center" alt="Diagram for puzzle" src="http://3.bp.blogspot.com/_oCutOBezbNs/TADy3vJidsI/AAAAAAAAAEE/v7UI2mUQfLo/s320/4ct+problem.jpg" border="0" /></a><br />You have to paint the marked regions in the diagram using four colours so that each region is a single colour and no regions with a common boundary are the same colour. Each region is 8 square metres in area except for the top area which is 16 square metres. You have only the following paint available: enough red for 24 square metres, enough yellow for 24 square metres, enough green for 16 square metres and enough blue for 8 square metres. Can you find a way to do this?<br /><br /><div></div>http://greenwich-maths-time.blogspot.com/2010/05/greenwich-maths-challenge-4.htmlnoreply@blogger.com (Tony)0tag:blogger.com,1999:blog-4236175346699186251.post-213172661337011673Mon, 24 May 2010 16:33:00 +00002010-05-29T11:50:22.603+01:00Greenwich Maths Challenge 3<a href="http://4.bp.blogspot.com/_oCutOBezbNs/S_qtSj25BXI/AAAAAAAAAD0/CH_Jeu98d8E/s1600/GMC3.jpg"><img id="BLOGGER_PHOTO_ID_5474878831396980082" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 151px; CURSOR: hand; HEIGHT: 67px; TEXT-ALIGN: center" alt="" src="http://4.bp.blogspot.com/_oCutOBezbNs/S_qtSj25BXI/AAAAAAAAAD0/CH_Jeu98d8E/s320/GMC3.jpg" border="0" /></a> The winner of <a href="http://greenwich-maths-time.blogspot.com/2010/02/greenwich-maths-challenge-3.html">the third Greenwich Maths Challenge</a> was Nic Mortimer, who was the first (and only) person to break the Vigenere cipher. The text was Hardy's famous reminiscence of Ramanujan,<br /><blockquote><p>I remember once going to see him when he was ill at Putney. I had ridden in taxi cab number **** and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavorable omen. "No," he replied, "it is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways."</p><p></p></blockquote><div>Here the missing number is of course 1729. The keyword to the cipher was "Hardy".</div><div></div><div>Greenwich Maths Challenge 4 will be posted here on or around Friday May 28, to mark the end of the exams at Greenwich! </div>http://greenwich-maths-time.blogspot.com/2010/05/greenwich-maths-challenge-3.htmlnoreply@blogger.com (Tony)0tag:blogger.com,1999:blog-4236175346699186251.post-317324072498162928Sun, 07 Feb 2010 09:30:00 +00002010-02-07T09:33:00.184+00:00Greenwich maths challenge cryptography cipherGreenwich Maths Challenge 3<a href="http://1.bp.blogspot.com/_oCutOBezbNs/S26IYsXy8ZI/AAAAAAAAADs/pWn6OH4m0Qw/s1600-h/GMC3.jpg"><img id="BLOGGER_PHOTO_ID_5435431758091973010" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 151px; CURSOR: hand; HEIGHT: 67px; TEXT-ALIGN: center" alt="CMC3 Logo" src="http://1.bp.blogspot.com/_oCutOBezbNs/S26IYsXy8ZI/AAAAAAAAADs/pWn6OH4m0Qw/s320/GMC3.jpg" border="0" /></a><br /><div>This was posted at 9:30am on Sunday 7 February but because I had prepared it in advance it doesn't appear as the newest item in the blog. <a href="http://greenwich-maths-time.blogspot.com/2010/02/greenwich-maths-challenge-3.html">Here is a direct link to it.</a></div>http://greenwich-maths-time.blogspot.com/2010/02/greenwich-maths-challenge-3_07.htmlnoreply@blogger.com (Tony)1tag:blogger.com,1999:blog-4236175346699186251.post-4914211713705395734Sun, 07 Feb 2010 09:00:00 +00002010-02-07T09:08:29.502+00:00Tomorrow's Mathematicians Today<a href="http://1.bp.blogspot.com/_oCutOBezbNs/S26Cmzy1xxI/AAAAAAAAADk/2Jqr5XDWfmw/s1600-h/Speakers-cropped.jpg"></a><br /><div><a href="http://2.bp.blogspot.com/_oCutOBezbNs/S26BV80GCxI/AAAAAAAAADc/oh5H0jmQ-oA/s1600-h/TMT-speakers-for-web.jpg"><img id="BLOGGER_PHOTO_ID_5435424014384630546" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 320px; CURSOR: hand; HEIGHT: 118px; TEXT-ALIGN: center" alt="The speakers at Tomorrow's Mathematicians Today" src="http://2.bp.blogspot.com/_oCutOBezbNs/S26BV80GCxI/AAAAAAAAADc/oh5H0jmQ-oA/s320/TMT-speakers-for-web.jpg" border="0" /></a><br />This was a wonderful conference with excellent presentations by undergraduates from all round the country, crowned by an exhilarating keynote address by Ian Stewart. The friendly atmosphere and evivent enjoyment of all the participants made it an extremely rewarding day.<br /><div></div></div>http://greenwich-maths-time.blogspot.com/2010/02/tomorrows-mathematicians-today.htmlnoreply@blogger.com (Tony)0tag:blogger.com,1999:blog-4236175346699186251.post-7107806498353222188Sat, 06 Feb 2010 20:38:00 +00002010-02-07T09:29:26.586+00:00Greenwich Maths Challenge 3<a href="http://2.bp.blogspot.com/_oCutOBezbNs/S23UrMmaTcI/AAAAAAAAADU/vtJrnJAMtDE/s1600-h/GMC3.jpg"><img id="BLOGGER_PHOTO_ID_5435234163887852994" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 151px; CURSOR: hand; HEIGHT: 67px; TEXT-ALIGN: center" alt="GMC3 logo" src="http://2.bp.blogspot.com/_oCutOBezbNs/S23UrMmaTcI/AAAAAAAAADU/vtJrnJAMtDE/s320/GMC3.jpg" border="0" /></a><br /><div><span style="font-family:times new roman;">The following text has been encrypted with a Vigenere cipher. (<a href="http://www.simonsingh.net/Code_Book_Download.html">Simon Singh's Code Book CD-rom</a>, which can be downloaded free, has a useful tool for deciphering such ciphers.) The text which has been enciphered contains one number which has been omitted. The prize for this challenge will be awarded to the first Greenwich maths undergraduate to email to <a href="mailto:A.Mann@gre.ac.uk">A.Mann@gre.ac.uk</a> two pieces of information: 1) the missing number and 2) the Vigenere keyword. </span></div><br /><div><span style="font-family:courier new;"></span></div><br /><div><span style="font-family:courier new;">PRVPCTBVUMUCVJMPNXWMZEVKGTWYHLOENDQPLCDRWUKQCFIYD<br />BYIUGCUIEWYEITDZUUDECYAEGPLMRUILDKKYATYHLBMSHPZEV<br />PCKTFPCYAKKCYAUXJSOEHYUDKKYAIYRNLDZWUHSERRHNLQDHV<br />FUYILVRKLNERFLRVSJPEULRPSRYCYYZQRLRVVRPNXQSTBVUGA<br />IJWFLSDDJSEJWLBMSHPLXGUCZSZEJLAJWFLSLPMMTNRABBVVG<br />UTNRBPFWHPLNKZYFS</span></div>http://greenwich-maths-time.blogspot.com/2010/02/greenwich-maths-challenge-3.htmlnoreply@blogger.com (Tony)0tag:blogger.com,1999:blog-4236175346699186251.post-2951416127357755301Sat, 23 Jan 2010 20:30:00 +00002010-01-30T16:28:47.252+00:00Greenwich Maths Challenge 3 - COMING 7 FEBRUARY<a href="http://3.bp.blogspot.com/_oCutOBezbNs/S1tcrEV1XCI/AAAAAAAAADM/VWB2tYjwzlU/s1600-h/GMC3.jpg"><img id="BLOGGER_PHOTO_ID_5430035670694714402" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 151px; CURSOR: hand; HEIGHT: 67px; TEXT-ALIGN: center" alt="GMC3 logo" src="http://3.bp.blogspot.com/_oCutOBezbNs/S1tcrEV1XCI/AAAAAAAAADM/VWB2tYjwzlU/s320/GMC3.jpg" border="0" /></a><br /><div>The third Greenwich Maths Challenge will be posted here on <strong>Sunday 7 February </strong>at <strong>9:30am</strong> (approximately). We have postponed it from the previously announced date so that those preparing presentations for the <a href="http://mathsoc.cms.gre.ac.uk/tmt">Tomorrow's Mathematicians Today conference</a> on 6 February will not be distracted! It will be a cryptography challenge so speed will be of the essence. If you don't already have it you might like to <a href="http://www.simonsingh.net/Code_Book_Download.html">download Simon Singh's free "Code Book" CD-rom</a> and practice cracking Vigenere ciphers!</div>http://greenwich-maths-time.blogspot.com/2010/01/greenwich-maths-challenge-3.htmlnoreply@blogger.com (Tony)0tag:blogger.com,1999:blog-4236175346699186251.post-4613644169645699720Sat, 23 Jan 2010 16:56:00 +00002010-01-23T20:26:22.031+00:00Christmas Quiz resultsCongratulations to Ameli Gottstein, who has (for the second year in a row) won our Christmas quiz. Here are the answers:<br /><br /><br />Q1. Identify the mathematicians whose names are given below as anagrams. Accents and punctuation marks such as hyphens are omitted, and spellings are taken from <a href="http://www-groups.dcs.st-and.ac.uk/~history/">the MacTutor History of Mathematics website</a><br /><br />a) ENRW - Wren<br /><br />b) EIKLNV - Kelvin<br /><br />c) EEKLPR - Kepler<br /><br />d) AAIIJNX - Jia Xian<br /><br />e) AHILMNOT - Hamilton<br /><br />f) BEILLNORU - Bernoulli<br /><br />g) ACDEEHIMRS - Archimedes<br /><br />h) AEMNNNNOUV - von Neumann<br /><br />i) GKLMOOOORV - Kolmogorov<br /><br />j) AAAEKKLOSVVY - Kovalevskaya<br /><br /><br />Q2. What is the smallest two-digit integer?<br /><br />-99. Silly, but it caught me out when I was asked.<br /><br /><br />Q3. Is 10^2010 + 1 prime? Justify your answer.<br /><br />The trick here is that for any integers x and y, x^3 - y^3 has a factor x-y. So if n is a multiple of 3, then 10^n+1 is the difference between two cubes because it is 10^(n/3)^3 - (-1)^3. So 10^2010+1 is not prime - in fact it has a factor 10^670+1.<br /><br /><br /><br />Q4. On average, how many times do you need to toss a fair coin before you have seen a run of an odd number of heads followed by a tail?<br /><br />This was from Peter Winkler, <em>Mathematical Puzzles: A Connoisseur's Collection</em> (A.K. Peters, 2004). Suppose that the answer is x. Consider the first couple of tosses. If the first coin comes up tails (which it will with probability 1/2), then we have made no progress and the expected number of tosses is now x+1. If the first coin comes up heads and the second toss is also heads (which happens one time in four), then we again expect to take a further x tosses so the expected number of tosses is now x+2. And if the first toss is heads and the second is tails (again one time in four), we have finished and it took two tosses. So we have<br />x = (1/2).(x+1) + (1/4).(x+2) + (1/4).2 so x = 6.<br /><br />Q5. <a href="http://2.bp.blogspot.com/_oCutOBezbNs/Synsyj-IGUI/AAAAAAAAADE/qbx0OiBCBa4/s1600-h/chessboard.jpg"></a>Consider a chessboard – an 8x8 grid of squares in which each row and column is alternately black and white. You have 31 2x1 rectangles each the size of two adjacent squares of the chessboard. Remove the top left and bottom right corner squares. Is it possible to cover the 62 remaining squares with your 31 rectangles? Provide a solution or show that it can’t be done.<br /><br /><br />This is a lovely old favourite. Colour the chessboard in the normal way, so that each row and each column contains alternating balck and white squares. Each of your rectangles, wherever it is placed, must cover one black and one white square. But the 62-square board we have to cover has 32 black and 30 white squares, so it can't be covered by our 31 pieces.<br /><br /><br />Q6. A, B and C are candidates in an election. There is an odd number of voters. The votes are counted and there is a three-way tie. As a tie-breaker the voters’ are asked for their second choices, and again there is a three-way tie. A suggests that, to break the tie, there be a two-way election between B and C, with the winner then facing A in another two-way election. Is this fair? And what is the probability that A would win the election if it were held in this way, assuming no voter changes their mind?<br /><br />I found this one in the same book by Peter Winkler as Q4: it was originally devised by Ehud Friedgut. The answer is that A's proposal is very far from fair: the probability that A would win under that system is 1. Suppose B beats C in the two-way election: that means that the majority of A's voters prefer B to C. But B has exactly one-third of the second preferences so, to balance, a minority of C's supporters must prefer B to A. Since C's supporters will decide the two-way election between A and B, A will win that election. A symmetrical argument shows that if C beats B in the two-way election then A will beat C in the run-off. So A is guaranteed to win.<br /><br /><br />Q7. Here are two hidden messages. What do they say?<br /><br />a) MERRY CHRISTMAS AND HAPPY NEW YEAR (this was originally written in invisible ink, ie in white text on a white background. Moving your mouse over it would reveal it!)<br /><br />b) AAAAA VHCTUMMLD<br /><br />No-one submitted a correct solution to this one. It uses a Vigenere cipher, with the "keyword" chosen so that the first word comes out as AAAAA. My intention was that likely guesses, from the word lengths with part (a) as a clue, would be "MERRY CHRISTMAS" and "HAPPY CHRISTMAS", and testing them with a Vigenere solver would reveal which one it was. In fact the keyword was "TALLC" and the message "Merry Christmas".http://greenwich-maths-time.blogspot.com/2010/01/christmas-quiz-results.htmlnoreply@blogger.com (Tony)0tag:blogger.com,1999:blog-4236175346699186251.post-4871304109254299029Thu, 14 Jan 2010 06:49:00 +00002010-01-14T06:51:07.764+00:00Christmas Quiz deadline extendedBecause of the bad weather the deadline for <a href="http://greenwich-maths-time.blogspot.com/2009/12/christmas-quiz.html">the Christmas Quiz</a> has been extended to 5pm on Monday 18 January.http://greenwich-maths-time.blogspot.com/2010/01/christmas-quiz-deadline-extended.htmlnoreply@blogger.com (Tony)0tag:blogger.com,1999:blog-4236175346699186251.post-4315348411360133992Thu, 17 Dec 2009 08:24:00 +00002010-01-14T06:53:44.527+00:00CHRISTMAS QUIZA small prize will be offered for the best solution emailed to A.Mann@gre.ac.uk by a Greenwich student before 5pm on Monday 18 January 2010 (deadline extended because of the bad weather). In the event of a tie, a winner will be chosen randomly. The judges' decision is final. For obvious reasons, the source of these questions won’t be revealed until afterwards. The quiz has seven questions.<br /><br />Q1. Identify the mathematicians whose names are given below as anagrams. Accents and punctuation marks such as hyphens are omitted, and spellings are taken from the MacTutor History of Mathematics website<br />http://www-groups.dcs.st-and.ac.uk/~history/<br /><br />a) ENRW<br />b) EIKLNV<br />c) EEKLPR<br />d) AAIIJNX<br />e) AHILMNOT<br />f) BEILLNORU<br />g) ACDEEHIMRS<br />h) AEMNNNNOUV<br />i) GKLMOOOORV<br />j) AAAEKKLOSVVY<br /><br /><br />Q2. What is the smallest two-digit integer?<br /><br />Q3. Is 10^2010 + 1 prime? Justify your answer.<br /><br />Q4. On average, how many times do you need to toss a fair coin before you have seen a run of an odd number of heads followed by a tail?<br /><br />Q5.<br /><a href="http://2.bp.blogspot.com/_oCutOBezbNs/Synsyj-IGUI/AAAAAAAAADE/qbx0OiBCBa4/s1600-h/chessboard.jpg"><img id="BLOGGER_PHOTO_ID_5416120380283164994" style="DISPLAY: block; MARGIN: 0px auto 10px; WIDTH: 206px; CURSOR: hand; HEIGHT: 169px; TEXT-ALIGN: center" alt="" src="http://2.bp.blogspot.com/_oCutOBezbNs/Synsyj-IGUI/AAAAAAAAADE/qbx0OiBCBa4/s320/chessboard.jpg" border="0" /></a> Consider a chessboard – an 8x8 grid of squares in which each row and column is alternately black and white. You have 31 2x1 rectangles each the size of two adjacent squares of the chessboard. Remove the top left and bottom right corner squares. Is it possible to cover the 62 remaining squares with your 31 rectangles? Provide a solution or show that it can’t be done.<br /><br />Q6. A, B and C are candidates in an election. There is an odd number of voters. The votes are counted and there is a three-way tie. As a tie-breaker the voters’ are asked for their second choices, and again there is a three-way tie. A suggests that, to break the tie, there be a two-way election between B and C, with the winner then facing A in another two-way election. Is this fair? And what is the probability that A would win the election if it were held in this way, assuming no voter changes their mind?<br /><br />Q7. Here are two hidden messages. What do they say?<br /><br />a) <span style="color:#ffffff;"><em>MERRY CHRISTMAS AND HAPPY NEW YEAR<br /></em></span><br />b) AAAAA VHCTUMMLDhttp://greenwich-maths-time.blogspot.com/2009/12/christmas-quiz.htmlnoreply@blogger.com (Tony)0